Detect and Remove Loop in a Linked List

Write a function detectAndRemoveLoop() that checks whether a given Linked List contains loop and if loop is present then removes the loop and returns true. And if the list doesn’t contain loop then returns false. Below diagram shows a linked list with a loop. detectAndRemoveLoop() must change the below list to 1->2->3->4->5->NULL.

Before trying to remove the loop, we must detect it. Techniques discussed in the above post can be used to detect loop. To remove loop, all we need to do is to get pointer to the last node of the loop. For example, node with value 5 in the above diagram. Once we have pointer to the last node, we can make the next of this node as NULL and loop is gone.
We can easily use Hashing or Visited node techniques (discussed in the above mentioned post) to get the pointer to the last node. Idea is simple: the very first node whose next is already visited (or hashed) is the last node.
We can also use Floyd Cycle Detection algorithm to detect and remove the loop. In the Floyd’s algo, the slow and fast pointers meet at a loop node. We can use this loop node to remove cycle. There are following two different ways of removing loop when Floyd’s algorithm is used for Loop detection.

Method 1 (Check one by one)
We know that Floyd’s Cycle detection algorithm terminates when fast and slow pointers meet at a common point. We also know that this common point is one of the loop nodes (2 or 3 or 4 or 5 in the above diagram). We store the address of this in a pointer variable say ptr2. Then we start from the head of the Linked List and check for nodes one by one if they are reachable from ptr2. When we find a node that is reachable, we know that this node is the starting node of the loop in Linked List and we can get pointer to the previous of this node.

Python

# Python program to detect and remove loop in linked list
# Node class
class Node:
# Constructor to initialize the node object
def __init__(self, data):
self.data = data
self.next = None
class LinkedList:
# Function to initialize head
def __init__(self):
self.head = None
def detectAndRemoveLoop(self):
slow_p = fast_p = self.head
while(slow_p and fast_p and fast_p.next):
slow_p = slow_p.next
fast_p = fast_p.next.next
# If slow_p and fast_p meet at some poin
# then there is a loop
if slow_p == fast_p:
self.removeLoop(slow_p)
# Return 1 to indicate that loop if found
return 1
# Return 0 to indicate that there is no loop
return 0
# Function to remove loop
# loop node-> Pointer to one of the loop nodes
# head --> Pointer to the start node of the
# linked list
def removeLoop(self, loop_node):
# Set a pointer to the beginning of the linked
# list and move it one by one to find the first
# node which is part of the linked list
ptr1 = self.head
while(1):
# Now start a pointer from loop_node and check
# if it ever reaches ptr2
ptr2 = loop_node
while(ptr2.next!= loop_node and ptr2.next !=ptr1):
ptr2 = ptr2.next
# If ptr2 reached ptr1 then there is a loop.
# So break the loop
if ptr2.next == ptr1 :
break
ptr1 = ptr1.next
# After the end of loop ptr2 is the lsat node of
# the loop. So make next of ptr2 as NULL
ptr2.next = None
# Function to insert a new node at the beginning
def push(self, new_data):
new_node = Node(new_data)
new_node.next = self.head
self.head = new_node
# Utility function to prit the linked LinkedList
def printList(self):
temp = self.head
while(temp):
print temp.data,
temp = temp.next
# Driver program
llist = LinkedList()
llist.push(10)
llist.push(4)
llist.push(15)
llist.push(20)
llist.push(50)
# Create a loop for testing
llist.head.next.next.next.next.next = llist.head.next.next
llist.detectAndRemoveLoop()
print "Linked List after removing loop"
llist.printList()
# This code is contributed by Nikhil Kumar Singh(nickzuck_007)

Output:

Linked List after removing loop
50 20 15 4 10

Method 2 (Better Solution)
This method is also dependent on Floyd’s Cycle detection algorithm.
1) Detect Loop using Floyd’s Cycle detection algo and get the pointer to a loop node.
2) Count the number of nodes in loop. Let the count be k.
3) Fix one pointer to the head and another to kth node from head.
4) Move both pointers at the same pace, they will meet at loop starting node.
5) Get pointer to the last node of loop and make next of it as NULL.

Method 3 (Optimized Method 2: Without Counting Nodes in Loop)
We do not need to count number of nodes in Loop. After detecting the loop, if we start slow pointer from head and move both slow and fast pointers at same speed until fast don’t meet, they would meet at the beginning of linked list.

How does this work?
Let slow and fast meet at some point after Floyd’s Cycle finding algorithm. Below diagram shows the situation when cycle is found.

We can conclude below from above diagram

Distance traveled by fast pointer = 2 * (Distance traveled
by slow pointer)
(m + n*x + k) = 2*(m + n*y + k)
Note that before meeting the point shown above, fast
was moving at twice speed.
x --> Number of complete cyclic rounds made by
fast pointer before they meet first time
y --> Number of complete cyclic rounds made by
slow pointer before they meet first time

From above equation, we can conclude below

m + k = (x-2y)*n
Which means m+k is a multiple of n.

So if we start moving both pointers again at same speed such that one pointer (say slow) begins from head node of linked list and other pointer (say fast) begins from meeting point. When slow pointer reaches beginning of linked list (has made m steps). Fast pointer would have made also moved m steps as they are now moving same pace. Since m+k is a multiple of n and fast starts from k, they would meet at the beginning. Can they meet before also? No because slow pointer enters the cycle first time after m steps.